Prove: If two circles are tangent internally at point P and the chords PA and (chord) PB of the larger circle intersect the smaller circle at points C and D respectively, then (chord) AB is parallel to (chord) CD.
What is your purpose in posting this? You show no attempt at all of your own to do the problem. If the two circles are "tangent internally" then the line perpendicular to the tangent line at P passes through the centers of both circles. That should make the problem easy.
http://mathhelpforum.com/newreply.ph...reply&p=921319
In your you assumed that $A~\&~B$ are on opposites sides of $\overleftrightarrow {PO}$ where $O$ is the center of one of the circles.
In the given they may well be on the same side. Look at my diagram.
The theorem that I'm using is this one -- https://en.wikipedia.org/wiki/Inscribed_angle. Yes obviously, if you read the proof, there is a typo. It should read triangle ABP is similar to triangle CDP.
As drawn, angle APB = angle CPD is acute. So $\angle AO_1B=2\,\angle APB=2\,\angle CPD=\angle CO_2D$. Similarly, if angle APB is obtuse. By the way, I realized this is not necessary to prove the similarity of triangles APB and CPD.
Here's a diagram where the angle at P is obtuse. The proof is the "same".